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Question Number 121539 by ajfour last updated on 09/Nov/20

Commented by ajfour last updated on 09/Nov/20

$$Find\frac{s}{R}formaximumblue$$$$triangulararea.$$$$\left(squareremainswithinsemicircle\right)$$

Answered by mr W last updated on 09/Nov/20

$$area\mathrm{\Delta }=\frac{sx}{2}$$$$(x+s)(2R-x-s)=s^2$$$${(x+s)}^2-2R(x+s)+s^2=0$$$$x+s=R-\sqrt{R^2-s^2}$$$$x=R-s-\sqrt{R^2-s^2}$$$$\mathrm{\Delta }=\frac{1}{2}(Rs-s^2-s\sqrt{R^2-s^2})$$$$\frac{d\mathrm{\Delta }}{ds}=0$$$$R-2s-\sqrt{R^2-s^2}+\frac{s^2}{\sqrt{R^2-s^2}}=0$$$$let\lambda =\frac{s}{R}$$$$\Rightarrow 1-2\lambda -\sqrt{1-{\lambda }^2}+\frac{{\lambda }^2}{\sqrt{1-{\lambda }^2}}=0$$$$\Rightarrow 1+\frac{{\lambda }^2}{\sqrt{1-{\lambda }^2}}=2\lambda +\sqrt{1-{\lambda }^2}$$$$\Rightarrow \lambda =\frac{s}{R}=0.8269$$

Commented by mr W last updated on 09/Nov/20

$$yessir,youareright!$$

Commented by ajfour last updated on 09/Nov/20

Thank you sir. i think you have checked when is the area a maximum for we get two positive values of lambda.

Commented by ajfour last updated on 09/Nov/20

$$\frac{2△}{R^2}=\lambda -{\lambda }^2+\lambda \sqrt{1-{\lambda }^2}$$$$\lambda =0.63111$$$$Sirthinkthisiscorrect.$$

Commented by ajfour last updated on 09/Nov/20

Answered by ajfour last updated on 09/Nov/20

Commented by ajfour last updated on 09/Nov/20

$$s=2R\sin \theta \cos \theta ;x=2R\cos {}^2\theta -s$$$$2△=sx$$$$let\frac{s}{R}=\lambda =\sin 2\theta$$$$2△=\left(R\sin 2\theta \right)(2R\cos {}^2\theta -R\sin 2\theta )$$$$\frac{2△}{R^2}=\sin 2\theta (1+\cos 2\theta -\sin 2\theta )$$$$\frac{d}{d\theta }\left(\frac{2△}{R^2}\right)=0\Rightarrow$$$$2\cos 2\theta (1+\cos 2\theta -\sin 2\theta )$$$$=\sin 2\theta (2\cos 2\theta +2\sin 2\theta )$$$$\Rightarrow \cos 2\theta +\cos {}^{2}2\theta =2\cos 2\theta \sin 2\theta$$$$+\sin {}^{2}2\theta$$$$with\sin 2\theta =\lambda$$$$(1-{\lambda }^2){(1-2\lambda )}^2={(2{\lambda }^2-1)}^2$$$$\Rightarrow 4{\lambda }^2-4{\lambda }^4-4\lambda +4{\lambda }^3+1-{\lambda }^2$$$$=4{\lambda }^4-4{\lambda }^2+1$$$$\Rightarrow 8{\lambda }^3-4{\lambda }^2-7\lambda +4=0$$$$\lambda =0.82694,0.63111$$$$Ithink\lambda =\frac{s}{R}=0.63111should$$$$betheappropriateanswer.$$$$$$

Answered by MJS_new last updated on 09/Nov/20

$$\mathrm{let}R=1$$$$\mathrm{semicircle}:y=\sqrt{1-x^2}$$$$s=\sqrt{1-p^2}$$$$\mathrm{base}\mathrm{of}\mathrm{triangle}\mid p-1+\sqrt{1-p^2}\mid$$$$2\times \mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mid p-1+\sqrt{1-p^2}\mid \sqrt{1-p^2}$$$$\mathrm{its}\mathrm{maximum}\mathrm{is}\mathrm{at}p\approx -.775694$$$$\Rightarrow s\approx .631109$$

Commented by MJS_new last updated on 09/Nov/20

$$\mathrm{please}\mathrm{check}...$$

Commented by ajfour last updated on 09/Nov/20

Commented by ajfour last updated on 09/Nov/20

$$s=\sqrt{1-p^2},R=1$$$$baseoftriangle\mathit{x}=1+p-\sqrt{1-p^2}$$$$2△=(1+p-\sqrt{1-p^2})\sqrt{1-p^2}$$$$=(1+p)\sqrt{1-p^2}-1+p^2$$$$\frac{d(2△)}{dp}=\sqrt{1-p^2}-\frac{p(1+p)}{\sqrt{1-p^2}}+2p=0$$$$\Rightarrow 1-p^2-p-p^2+2p\sqrt{1-p^2}=0$$$$\Rightarrow {(2p^2+p-1)}^2=4p^2(1-p^2)$$$$\Rightarrow 8p^4+4p^3-7p^2-2p+1=0$$$$max(2△)\approx 0.72236$$$$atp\approx 0.77569$$$$s=0.63111$$$$IfthisiswhatyoumeanSirMjS.$$

Commented by ajfour last updated on 09/Nov/20

Commented by MJS_new last updated on 09/Nov/20

$$\mathrm{yes}\mathrm{thank}\mathrm{you},\mathrm{typo}\mathrm{while}\mathrm{approximating}\mathrm{the}$$$$\mathrm{zero}.\mathrm{I}\mathrm{have}\mathrm{corrected}\mathrm{it}$$

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